dạ đây 

Ta có $\dfrac{{{x^2}}}{{2{x^2} - 2x + 1}} = \dfrac{{{x^2}}}{{{x^2} + {{(x - 1)}^2}}}$

$\begin{array}{l}
 \Leftrightarrow S = f(2016) + f( - 2015) + f(2015) + f( - 2014)... + f(2) + f( - 1) + f(1) + f(2017) + f( - 2016)\\
 = \dfrac{{{{2016}^2}}}{{{{2016}^2} + {{2015}^2}}} + \dfrac{{{{2015}^2}}}{{{{2015}^2} + {{2016}^2}}} + \dfrac{{{{2015}^2}}}{{{{2015}^2} + {{2014}^2}}} + \dfrac{{{{2014}^2}}}{{{{2014}^2} + {{2013}^2}}} + ... + \dfrac{{{2^2}}}{{{2^2} + {1^2}}} + \dfrac{{{1^2}}}{{{1^2} + {2^2}}} + \dfrac{{{1^2}}}{{{1^2} + {0^2}}} + \dfrac{{{{2017}^2}}}{{{{2017}^2} + {{2016}^2}}} + \dfrac{{{{2016}^2}}}{{{{2016}^2} + {{2017}^2}}}\\
 = \dfrac{{{{2016}^2} + {{2015}^2}}}{{{{2016}^2} + {{2015}^2}}} + \dfrac{{{{2015}^2} + {{2014}^2}}}{{{{2015}^2} + {{2014}^2}}} + ... + \dfrac{{{2^2} + {1^2}}}{{{2^2} + {1^2}}} + 1 + \dfrac{{{{2017}^2} + {{2016}^2}}}{{{{2017}^2} + {{2016}^2}}}\\
 = 1 + 1 + ... + 1 + 1 + 1 = 2017
\end{array}$